1053 Path of Equal Weight (30分)--树的DFS

https://pintia.cn/problem-sets/994805342720868352/problems/994805424153280512

1053 Path of Equal Weight (30分)

Given a non-empty tree with root R, and with weight W​i​assigned to each tree node T​i​​ . The weight of a path from R to L is defined to be the sum of the weights of all the nodes along the path from R to any leaf node L.

Now given any weighted tree, you are supposed to find all the paths with their weights equal to a given number. For example, let’s consider the tree showed in the following figure: for each node, the upper number is the node ID which is a two-digit number, and the lower number is the weight of that node. Suppose that the given number is 24, then there exists 4 different paths which have the same given weight: {10 5 2 7}, {10 4 10}, {10 3 3 6 2} and {10 3 3 6 2}, which correspond to the red edges in the figure.

Input Specification:
Each input file contains one test case. Each case starts with a line containing 0<N≤100, the number of nodes in a tree, M (<N), the number of non-leaf nodes, and 0<S<2³⁰​​ , the given weight number. The next line contains N positive numbers where W​i​​ (<1000) corresponds to the tree node T​i​​ . Then M lines follow, each in the format:

ID K ID[1] ID[2] … ID[K]
where ID is a two-digit number representing a given non-leaf node, K is the number of its children, followed by a sequence of two-digit ID’s of its children. For the sake of simplicity, let us fix the root ID to be 00.

Output Specification:
For each test case, print all the paths with weight S in non-increasing order. Each path occupies a line with printed weights from the root to the leaf in order. All the numbers must be separated by a space with no extra space at the end of the line.

Note: sequence {A​1​​ ,A​2​​ ,⋯,A​n​​ } is said to be greater than sequence {B​1​​ ,B​2​​ ,⋯,B​m​​ } if there exists 1≤k<min{n,m} such that A​i​​ =B​i​​ for i=1,⋯,k, and Ak+1​ >B​k+1​​ .

Sample Input:
20 9 24
10 2 4 3 5 10 2 18 9 7 2 2 1 3 12 1 8 6 2 2
00 4 01 02 03 04
02 1 05
04 2 06 07
03 3 11 12 13
06 1 09
07 2 08 10
16 1 15
13 3 14 16 17
17 2 18 19
Sample Output:
10 5 2 7
10 4 10
10 3 3 6 2
10 3 3 6 2

#include<bits/stdc++.h>
using namespace std;
struct Node{
	int w;
	vector<int>child;
};
vector<Node>node;
vector<int>path;
int n,m,S;
bool cmp(int a,int b){
	return node[a].w>node[b].w;
}
void DFS(int index,int numNode,int numSum){
	if(numSum>S)return ;
	if(S==numSum){
		if(node[index].child.size()!=0)return ;
		for(int i=0;i<numNode;i++){
			printf("%d%c",node[path[i]].w,i!=numNode-1 ?' ':'\n');
		}
	}
	for(int i=0;i<node[index].child.size();i++){
		int child=node[index].child[i];
		path[numNode]=child;
		DFS(child,numNode+1,numSum+node[child].w);
	}
}
int main(){
	scanf("%d%d%d",&n,&m,&S);
	node.resize(n),path.resize(n);
	for(int i=0;i<n;i++){
		scanf("%d",&node[i].w);
	}
	int id,child,k;
	for(int i=0;i<m;i++){
		scanf("%d%d",&id,&k);
		for(int j=0;j<k;j++){
			scanf("%d",&child);
			node[id].child.push_back(child);
		}
		sort(node[id].child.begin(),node[id].child.end(),cmp);
	} 
	DFS(0,1,node[0].w);
	return 0;
}